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Let $a_{n+1}= \sqrt{a_n +1}$ for all natural numbers $n$.

How do you prove $a_n$ is contractive?

Definition of contractive sequence is:

If there exists a constant $C \in (0,1)$ such that $|a_{n+2} - a_{n+1}| \leq C|a_{n+1}-a_n|$ for all natural numbers $n$, then the sequence $a_n$ is contractive.

Here is my thought process:

First, I prove $a_n$ is monotone increasing by induction.
Second, I prove it is bounded above. Third, I compute what the limit would be.

Last, I try to get the required inequality for contrative sequence. I am stuck on this last step.

In the third step, I solved the quadratic equation $L^2 - L - 1 = 0$, which gives me L = $\frac{1+\sqrt{5}}{2}$

Then,for the fourth step, I do $|a_{n+2} - a_{n+1}|=|\sqrt{a_{n+1}+1} -\sqrt{a_{n}+1} | \leq L + L$. At this point I am not clear how to find the value of $C$.

user1068636
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2 Answers2

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Hint: $\;\left|a_{n+2} - a_{n+1}\right|=\left|\sqrt{a_{n+1}+1} -\sqrt{a_{n}+1}\right| = \dfrac{\left|a_{n+1}-a_n\right|}{\sqrt{a_{n+1}+1} + \sqrt{a_{n}+1}} \le \dfrac{\left|a_{n+1}-a_n\right|}{1+1}\,$.

dxiv
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Let $f(x)=\sqrt {1+x}\;$ for $x> 0.$ Then $f'(x)=\frac {1}{2\sqrt {1+x}}.$ So $0<f'(x)< \frac {1}{2}$ when $x>0.$

If $x,y$ are positive and $x> y$ then by the MVT there exists $z \in (y,z)$ such that $\frac {f(x)-f(y)}{x-y}=f'(z).$ Since $0<f'(z)<\frac {1}{2},$ therefore $0<f(x)-f(y)<\frac {1}{2}(x-y).$

Hence for any positive $x,y$ we have $|f(x)-f(y)|\leq \frac {1}{2}|x-y|$, with equality iff $x=y$.

Of course proving the validity of the formula for $f'(x)$ can be done by using the method in the Answer by dxiv.

DanielWainfleet
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